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third and sixth to the fourth. The demonstration of this is the same with that of the proposition, if division be used instead of composition.

COR. II. The proposition holds true of two ranks of magnitudes, whatever be their number, of which each of the first rank has to the second magnitude the same ratio that the corresponding one of the second rank has to a fourth magnitude; as is manifest.

THE

ELEMENT

OF

GEOMETRY.

BOOK V.

DEFINITIONS.

I.

Two magnitudes may be said to be reciprocally proportional to two others, when one of the first is to one of the second, as the remaining one of the second is to the remaining one of the first.

II.

A straight line may be said to be cut in extreme and mean ratio, when the whole is to the greater segment, as the greater segment is to the less.

III.

When two magnitudes can be divided into any number of equal parts, so that a part of one is equal to a part of the other, the magnitudes may be said to be commeasurable.

IV.

When they cannot be so divided, they may be said to be incommeasurable.

V.

Rectilineal figures which have their several angles equal, each to each, and the sides about the equal angles proportional, differ only in size, and therefore are the same in shape or similar.

PROP. I. 'PROBLEM.

To find the greatest common measure of two given straight lines.

Let AB and CD be the two given straight lines, it is required to find their greatest common measure. Apply the less CD to the.

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greater, as many times as it can be contained therein, for instance, twice, with the remainder BE; apply the remainder BE in the same way to CD, for instance, once, with the second remainder DF; apply DF in the same way to BE, for instance, once, with the third remainder BG; apply the third remainder BG to the second remainder DF; and continue thus until a remainder is found which is contained in the preceding without leaving a remainder.

Because the less line CD is to be measured, the common measure cannot be greater than CD, and if CD measures AB without a remainder, it is the greatest common measure. But if a remainder is left as BE, then the greatest common measure is less than CD, and not greater than BE; and if BE measures CD without a remainder, BE is the greatest common measure; and it may be shewn in the same way, that the first remainder that measures the preceding without a remainder, is the greatest common measure.

PROP. II. THEOR.

Parallelograms of the same altitude are to one another as their bases.

D

K

F

C

Let ABCD and AEFD, be two parallelograms which have the same altitude, they are to each other as their bases AB, AE. If AB and AE are commeasurable, let AG be the common measure; divide AB and AE into the equal parts AG, GE, and EB, and draw GK, EF, parallel to AD or BC (4. 2.); then the parallelograms thus made will be equal, and the parallelograms ABCD and AEFD will be to each other as the respective numbers of the equal parallelograms of which they are composed; and because the bases

A

G

E

B

have been divided into equal parts, these numbers will be to each other as the bases; and therefore the parallelograms ABCD, AEFD, will be to each other as the bases.

But if the bases AB and AE are not commeasurable, still ABCD : AEFD AB AE.

:

D.

For if not, let ABCD :AEFD: AB: AI, longer than AE; divide AB into equal parts less than EI, and let one of the divisions fall between E and I at K, and draw KL parallel to BC: then AB and AK being commeasurable, A

FL

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ABCD AKLD :: AB: AK; but by the hypothesis, ABCD: AEFD :: AB : AI; and in these two proportions the antecedents being equal, the consequents are proportional (13. 4.); and therefore, AKLD: AEFD:: AK: AI, but AI is longer than AK, and therefore AEFD must be larger than AKLD, whereas it is less; and therefore ABCD is not to AEFD as AB to AI, any line longer than AE. It may be shewn in the same way, that ABCD is not to AEFD as AB to any line shorter than AE; and therefore, ABCD: AEFD :: AB: AE. Therefore, Parallelograms, &c. Q. E. D.

COR. I. Because parallelograms of equal base and altitude are equal, therefore parallelograms of equal altitude are to one another as their bases.

COR. II. Because a triangle is equal to half a parallelogram of equal base and altitude, therefore triangles of equal altitude are to one another as their bases.

PROP. III. THEOR.

If a straight line be drawn parallel to one of the sides of a triangle, it shall cut the other sides, or these produced, proportionally; and if the sides, or the sides produced be cut proportionally, the straight line which joins the points of section shall be parallel to the remaining side of the triangle.

Let DE be drawn parallel to BC, one of the sides of the triangle ABC; BD is to DA, as CE to EA.

Join BE, CD; then the triangle BDE is equal to the triangle CDE (Cor. 1. 15. 2.), because they are on the same base DE, and between the same parallels DE, BC. ADE is another triangle, and equal magnitudes have to the same, the same ratio (13. 4.); therefore as the triangle BDE to the triangle ADE, so is the triangle CDE to the triangle ADE; but as the triangle BDE to the triangle ADE, so is (Cor. 2. 2. 5.) BD to DA, because having the same altitude, viz. the perpendicular drawn from the point E to AB, they are to one another as their bases; and for the same reason, as the triangle CDE to the triangle ADE, so is CE to EA: therefore as BD to DA, so is CE to EA.

Next, let the sides AB, AC of the triangle ABC, or these produced,

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be cut proportionally in the points D, E, that is, so that BD be to DA, as CE to EA, and join DE; DE is parallel to BC.

The same construction being made; because as BD to DA, şo is CE to EA; and as BD to DA, so is the triangle BDE to the triangle ADE (2. Cor. 2. 5.); and as CE to EA, so is the triangle CDE to the triangle ADE; therefore the triangle BDE is to the triangle ADE, as the triangle CDE to the triangle ADE; that is, the triangles BDE, CDE have the same ratio to the triangle ADE; and therefore (14. 4.) the triangle BDE is equal to the triangle CDE; and they are on the same base DE; but equal triangles on the same base are between the same parallels (2. Cor. 15. 2.), therefore DE is' parallel to BC. Wherefore, if a straight line, &c. Q. E. D.

PROP. IV. THEOR.

If the angle of a triangle be divided into two equal angles, by a straight line which also cuts the base; the segments of the base shall have the same ratio which the other sides of the triangle have to one another and if the segments of the base have the same ratio which the other sides of the triangle have to one another, the straight line drawn from the vertex to the point of section, divides the vertical angle into two equal angles.

Let the angle BAC of any triangle ABC be divided into two equal angles by the straight line AD: BD is to DC, as BA to AC. Through the point C draw CE

parallel (4. 2.) to DA, and let BA produced meet CE in E.

Because the straight line AC meets the parallels AD, EC, the angle ACE is equal to the corresponding angle CAD (3. 2.); but CAD by the hypothesis is equal to the angle BAD; wherefore BAD is equal to the angle ACE. Again, because the straight line BAE meets the parallels AD, EC, the outward angle BAD is

B

D

E

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